Python 基于2个数据帧的列生成

我有一个数据帧df1，它有：

F_Id    I_Code  F_Date
FT-56832    2   01/09/2019
FT-93828    1   01/09/2019
FT-13853    2   02/09/2019
FT-18858    3   02/09/2019
FT-19010    2   03/09/2019
FT-62064    5   02/09/2019
FT-94494    4   03/09/2019
FT-73594    2   03/09/2019
FT-78590    3   01/09/2019
FT-14296    4   01/09/2019
FT-82529    3   03/09/2019
FT-33266    3   04/09/2019
FT-58456    4   02/09/2019
FT-16693    4   04/09/2019
FT-69073    4   02/09/2019
FT-69649    1   05/09/2019

每个（I_代码、F_日期）都有5个不同的ID与之关联

我有另一个dataframe df2，它有以下列：

F_Date  num_i_found
01/09/2019  5
01/09/2019  3
02/09/2019  5
02/09/2019  5
03/09/2019  3
02/09/2019  4
03/09/2019  4
03/09/2019  5
01/09/2019  5
01/09/2019  4
03/09/2019  3
04/09/2019  5
02/09/2019  4
04/09/2019  5
02/09/2019  4
05/09/2019  4

我想生成一个在df2中找到的新列ID_，这样它就是一个ID为的数组

例如，对于2019年9月1日，num_i_found是4，那么ID_found将是df1中5个ID中的4个。（FT-56832，FT-93828，F-78590，…）

---

有没有办法达到同样的效果

创建列表字典，并按

num\u i\u found

值编制索引进行筛选：

注意：如果值与第一行不匹配，那么样本数据中的值仅为

2019年9月1日
的
4
值，我猜在实际数据中，每个日期时间都是
d
中的5个值，所以按照您的需要工作

d = df1.groupby('F_Date')['F_Id'].apply(list).to_dict()

print (d)
{'01/09/2019': ['FT-56832', 'FT-93828', 'FT-78590', 'FT-14296'],
 '02/09/2019': ['FT-13853', 'FT-18858', 'FT-62064', 'FT-58456', 'FT-69073'],
 '03/09/2019': ['FT-19010', 'FT-94494', 'FT-73594', 'FT-82529'], 
 '04/09/2019': ['FT-33266', 'FT-16693'], 
 '05/09/2019': ['FT-69649']}


如果需要字符串：

d = df1.groupby('F_Date')['F_Id'].apply(list).to_dict()

df2['new'] = df2.apply(lambda x: ', '.join(d.get(x['F_Date'], [])[:x['num_i_found']]), axis=1)
print (df2)
        F_Date  num_i_found                                               new
0   01/09/2019            5            FT-56832, FT-93828, FT-78590, FT-14296
1   01/09/2019            3                      FT-56832, FT-93828, FT-78590
2   02/09/2019            5  FT-13853, FT-18858, FT-62064, FT-58456, FT-69073
3   02/09/2019            5  FT-13853, FT-18858, FT-62064, FT-58456, FT-69073
4   03/09/2019            3                      FT-19010, FT-94494, FT-73594
5   02/09/2019            4            FT-13853, FT-18858, FT-62064, FT-58456
6   03/09/2019            4            FT-19010, FT-94494, FT-73594, FT-82529
7   03/09/2019            5            FT-19010, FT-94494, FT-73594, FT-82529
8   01/09/2019            5            FT-56832, FT-93828, FT-78590, FT-14296
9   01/09/2019            4            FT-56832, FT-93828, FT-78590, FT-14296
10  03/09/2019            3                      FT-19010, FT-94494, FT-73594
11  04/09/2019            5                                FT-33266, FT-16693
12  02/09/2019            4            FT-13853, FT-18858, FT-62064, FT-58456
13  04/09/2019            5                                FT-33266, FT-16693
14  02/09/2019            4            FT-13853, FT-18858, FT-62064, FT-58456
15  05/09/2019            4                                          FT-69649

我想在df2中获得新专栏。i、 e.F_date，num_i_found，new如果num_i_found是5，那么从df1中选择所有ID并追加到数组中并添加到新的column@user3759616-oops，我将其分配给df1，而不是df2，这是不正确的。与第一个条目（即2019年9月1日）一样，截至数据共享，df1中有4个条目，但实际上有5个，这会选择5，然后分配给列'new'@user3759616-刚刚添加的通知；）
d = df1.groupby('F_Date')['F_Id'].apply(list).to_dict()

df2['new'] = df2.apply(lambda x: ', '.join(d.get(x['F_Date'], [])[:x['num_i_found']]), axis=1)
print (df2)
        F_Date  num_i_found                                               new
0   01/09/2019            5            FT-56832, FT-93828, FT-78590, FT-14296
1   01/09/2019            3                      FT-56832, FT-93828, FT-78590
2   02/09/2019            5  FT-13853, FT-18858, FT-62064, FT-58456, FT-69073
3   02/09/2019            5  FT-13853, FT-18858, FT-62064, FT-58456, FT-69073
4   03/09/2019            3                      FT-19010, FT-94494, FT-73594
5   02/09/2019            4            FT-13853, FT-18858, FT-62064, FT-58456
6   03/09/2019            4            FT-19010, FT-94494, FT-73594, FT-82529
7   03/09/2019            5            FT-19010, FT-94494, FT-73594, FT-82529
8   01/09/2019            5            FT-56832, FT-93828, FT-78590, FT-14296
9   01/09/2019            4            FT-56832, FT-93828, FT-78590, FT-14296
10  03/09/2019            3                      FT-19010, FT-94494, FT-73594
11  04/09/2019            5                                FT-33266, FT-16693
12  02/09/2019            4            FT-13853, FT-18858, FT-62064, FT-58456
13  04/09/2019            5                                FT-33266, FT-16693
14  02/09/2019            4            FT-13853, FT-18858, FT-62064, FT-58456
15  05/09/2019            4                                          FT-69649